If the normal to the ellipse x^2 + 3y^2 = 12 at the point be inclined at 60° to the major axis, then at what angle does the line joining the curve to the point is inclined to the same axis?
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My question is taken from Calculus Application topic in section Application of Calculus of Mathematics – Class 12
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Right answer is (d) 30°
The best I can explain: Given, x^2 + 3y^2 = 12 Or x^2/12 + y^2/4 = 1
Differentiating both sides of (1) with respect to y we get,
2x*(dx/dy) + 3*2y = 0
Or dx/dy = -3y/x
Suppose the normal to the ellipse (1) at the point P(√12cosθ, 2sinθ) makes an angle 60° with the major axis. Then, the slope of the normal at P is tan60°
Or -[dx/dy]P = tan60°
Or -(-(3*2sinθ)/√12cosθ) = √3
Or √3tanθ = √3
Or tanθ = 1
Now the centre of the ellipse (1) is C(0, 0)
Therefore, the slope of the line CP is,
(2sinθ – 0)/(√12cosθ – 0) = (1/√3)tanθ = 1/√3 [as, tanθ = 1]
Therefore, the line CP is inclined at 30° to the major axis.